Rate of Change

[nycmathguy]

Exercises 1.1
5 & 6

Please, do number 6 showing a step by step solution. I will then do 5 and show my work as well.

Note: Not one section, not one chapter will be skipped throughout the textbook. This Ron Larson & Bruce Edwards textbook will be covered thoroughly.

 

[MathLover1]

5. A bicyclist is riding on a path modeled by function f(x)=0.04(8x-x^2), where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x=2.

Rate of change of a function is the first derivative of the function f(x).

so, you need to find derivate f'(x) of given function (which is actually slope of secant line)

f'(x)=0.04(1*8x^0-2*x^1)
f'(x)=0.04(8-2x) ........... plug in x=2
f'(2)=0.04(8-2*2)
f'(2)=0.04(4)
f'(2)=0.16

Rate of change of elevation at 2 is 0.16

6. A bicyclist is riding on a path modeled by function f(x)=0.08x, where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x=2.

figure shows graph of f(x)=0.08x which is linear function which has a constant slope of 0.08 (rate of change)
the rate of change of elevation is constant, = 0.08 for all values of x.
or, using derivate
f'(x)=1*0.08x^0
f'(x)=0.08*1
f'(x)=0.08
plug in x=2
f'(2)=0.08
Rate of change of elevation at 2 is 0.08

 

[nycmathguy]

5. A bicyclist is riding on a path modeled by function f(x)=0.04(8x-x^2), where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x=2.

Rate of change of a function is the first derivative of the function f(x).

so, you need to find derivate f'(x) of given function (which is actually slope of secant line)

f'(x)=0.04(1*8x^0-2*x^1)
f'(x)=0.04(8-2x) ........... plug in x=2
f'(2)=0.04(8-2*2)
f'(2)=0.04(4)
f'(2)=0.16

Rate of change of elevation at 2 is 0.16

6. A bicyclist is riding on a path modeled by function f(x)=0.08x, where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x=2.

figure shows graph of f(x)=0.08x which is linear function which has a constant slope of 0.08 (rate of change)
the rate of change of elevation is constant, = 0.08 for all values of x.
or, using derivate
f'(x)=1*0.08x^0
f'(x)=0.08*1
f'(x)=0.08
plug in x=2
f'(2)=0.08
Rate of change of elevation at 2 is 0.08

You are amazingly smart.

 

[nycmathguy]

5. A bicyclist is riding on a path modeled by function f(x)=0.04(8x-x^2), where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x=2.

Rate of change of a function is the first derivative of the function f(x).

so, you need to find derivate f'(x) of given function (which is actually slope of secant line)

f'(x)=0.04(1*8x^0-2*x^1)
f'(x)=0.04(8-2x) ........... plug in x=2
f'(2)=0.04(8-2*2)
f'(2)=0.04(4)
f'(2)=0.16

Rate of change of elevation at 2 is 0.16

6. A bicyclist is riding on a path modeled by function f(x)=0.08x, where x and f(x) are measured in miles (see figure). Find the rate of change of elevation at x=2.

figure shows graph of f(x)=0.08x which is linear function which has a constant slope of 0.08 (rate of change)
the rate of change of elevation is constant, = 0.08 for all values of x.
or, using derivate
f'(x)=1*0.08x^0
f'(x)=0.08*1
f'(x)=0.08
plug in x=2
f'(2)=0.08
Rate of change of elevation at 2 is 0.08

I thank you for showing your work but read my instructions again:

Please, do number 6 showing a step by step solution. I will then do 5 and show my work as well.

I wanted to do 5 as a test for me. You see, if I cannot solve the problems using your reply, then something is wrong.